Abel's test for uniform convergence in Fourier series In Abel's test for uniform convergence, we wite the terms of the series as $u_n(x)=a_n f_n(x)$ and there is a condition which says that the functions $f_n(x)$ should be monotonic, in mathematical notation it is 
$$
f_{n+1}(x) \leq f_n(x)
$$
Up to this point, things are fine, but when I am trying to look for uniform convergence in Fourier series I am unable to satisfy this condition. As an example in the series
$$
\sum_{n=1}^{\infty} \frac{\cos(nx)}{n^2}
$$
$f_n(x)=\cos(nx)$ which are not monotonic.
clear from this plot
Are the conditions for Abel's test are different for series that contain sinusoidal terms? 
Ref: Arfken 6th Edition page 351
 A: To just consider the question of uniform convergence with regard to your example, note that
$$\left|\frac{\cos nx}{n^2} \right| \leqslant \frac{1}{n^2}$$
Therefore, the series is absolutely and uniformly convergent for all $x \in \mathbb{R}$ by the Weierstrass M-test. Nevertheless, Abel’s test is not applicable.
Consider a similar example $\sum \frac{\cos nx}{n}$, which actually is conditionally convergent and the Weierstrass test cannot be applied.  Abel's test is not helpful since two conditions are violated in that $\cos nx$ is not monotonic and $\sum \frac{1}{n}$ is not convergent. Furthermore, even though $\frac{1}{n}$ is monotonic, the series $\sum \cos nx $ is not uniformly (nor even pointwise) convergent.
The series is not uniformly convergent for all $x \in \mathbb{R}$, but we can show that convergence is uniform on intervals that do not have $2k\pi$ for $k \in \mathbb{Z}$ as a limit point. To this end, we can use the related Dirichlet's test.
Consider, for example, the interval $[a,\pi]$ where $0 < a < \pi$. We see that $\frac{1}{n}$ is nonincreasing and uniformly convergent to $0$ as $n \to \infty$. Additionally, we have
$$\left|\sum_{n=1}^m \cos nx  \right| =  \left|\frac{\sin \frac{mx}{2} \cos \frac{(m+1)x}{2} }{\sin \frac{x}{2}}\right| \leqslant \frac{1}{\sin \frac{a}{2} }$$
Since these partial sums are uniformly bounded for all $m \in \mathbb{N}$ and $x \in [a,\pi]$ we have uniform convergence.
